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Ta có:
\(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+....+\left(1-\frac{1}{2016.2017}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\right)\)
\(=2016-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2016-\left(1-\frac{1}{2017}\right)\)
\(=2016-\frac{2016}{2017}=\frac{4064256}{2017}\)
Vậy giá trị biểu thức là \(\frac{4064256}{2017}\)
Biểu thức 1 :
1 4/5 x 5/6 = 9/5 x 5/6 = 45 /30 = 3/2
Biểu thức 2 :
15/16 : 11 /4 = 15/16 x 4/11 = 15/44
Tk cho mk nha ! Kết bạn với mk nhé !
\(\frac{1}{6.10}\)+ \(\frac{1}{10.14}\)+ ... + \(\frac{1}{402.406}\)
= \(\frac{1}{4}\). \(\left(\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\right)\)
= \(\frac{1}{4}\). ( \(\frac{10-6}{6.10}\)+ \(\frac{14-10}{10.14}\)+ ... + \(\frac{406-402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{10}{6.10}\)- \(\frac{6}{6.10}\)+ ... + \(\frac{406}{402.406}\)- \(\frac{402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{1}{6}\)- \(\frac{1}{406}\))
= \(\frac{1}{4}\). \(\frac{100}{609}\)
= \(\frac{25}{609}\)
\(\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)
\(=\frac{\frac{3.2}{2}}{2}+\frac{\frac{4.3}{2}}{3}+...+\frac{\frac{200.199}{2}}{199}\)
\(=\frac{3.2}{2}.\frac{1}{2}+\frac{4.3}{2}.\frac{1}{3}+...+\frac{200.199}{2}.\frac{1}{199}\)
\(=\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)
\(=\frac{3+4+...+200}{2}=\frac{203.198}{2}.\frac{1}{2}=\frac{20097}{2}\)
=17/6+1/2×13/3-9/4:3/5
=17/6+13/6-9/4×5/3
=17/6+13/6-15/4
=68+52-90/24
=30/24
=5/4
\(2\frac{5}{6}+0,5\times4\frac{1}{3}-2\frac{1}{4}:\frac{3}{5}\)
\(=\frac{17}{6}+\frac{1}{2}\times\frac{13}{3}-\frac{9}{4}:\frac{3}{5}\)
\(=\frac{17}{6}+\frac{13}{6}-\frac{9}{4}\times\frac{5}{3}\)
\(=5-\frac{9\times5}{4\times3}\)
\(=5-\frac{3\times5}{4\times1}\)
\(=5-\frac{15}{4}=\frac{20-15}{4}=\frac{5}{4}\)
1/1+2+1/1+2+3+1/1+2+3+4+1/1+2+3+4+5
=1/3+1/6+1/10+1/15
lấy mẫu số chung là 30,ta có:
10/30+5/6+3/10+2/30
=20/30=2/3
\(\frac{2016,2017-198}{2017.2015+1819}=\frac{\left(2015+1\right).2017-198}{2017.2015+1819}\)
\(=\frac{2015.2017+2017-198}{2017.2015+1819}\)
\(=\frac{2015.2017+1819}{2017.2015+1819}\)
\(=1\)
Nhớ k cho mình nha
\(\frac{2016.2017-198}{2017.2015+1819}=\frac{\left(2015+1\right)2017-198}{2017.2015+1819}=\frac{2015.2017+2017-198}{2017.2015+1819}=\frac{2017.2015+1819}{2017.2015+1819}=1\)